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Magazine Chapter 1: Q6E (page 1)
Question: Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the normal distribution with unknown mean \({\bf{\mu }}\) and unknown variance \({{\bf{\sigma }}^{\bf{2}}}\) . Let \(\widehat {{\bf{\sigma }}_{\bf{0}}^{\bf{2}}}\) and \(\widehat {{\bf{\sigma }}_{\bf{1}}^{\bf{2}}}\) be the two estimators of \({{\bf{\sigma }}^{\bf{2}}}\), which are defined as follows:
\(\widehat {{\bf{\sigma }}_{\bf{0}}^{\bf{2}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}{\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}\overline {\bf{X}} } \right)} ^{\bf{2}}}\)and \(\widehat {{\bf{\sigma }}_{\bf{1}}^{\bf{2}}}{\bf{ = }}\frac{{\bf{1}}}{{{\bf{n - 1}}}}{\sum\limits_{{\bf{i = 1}}}^{\bf{n}} {\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}\overline {\bf{X}} } \right)} ^{\bf{2}}}\)Show that the M.S.E. of \(\widehat {{\bf{\sigma }}_{\bf{0}}^{\bf{2}}}\) is smaller than the M.S.E. of \(\widehat {{\bf{\sigma }}_{\bf{1}}^{\bf{2}}}\) for all possible values of \({\bf{\mu }}\) and \({{\bf{\sigma }}^{\bf{2}}}\) .
Short Answer
M.S.E. of \(\widehat {\sigma _0^2}\) is smaller than the M.S.E. of \(\widehat {\sigma _1^2}\) for all possible values of \(\mu \) and \({\sigma ^2}\)
Step by step solution
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